Ch 19 PPT.ppt

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First Law of Thermodynamics

• Energy can be neither created nor

destroyed.Therefore, the total energy of the universe is a constant.

• Energy can, however, be converted

from one form to another or transferred from a system to the surroundings or vice versa.

Spontaneous Processes

• Spontaneous processes are those that can

proceed without any outside intervention.

• The gas in vessel B will

spontaneously effuse into

vessel A, but once the

gas is in both vessels, it

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Spontaneous Processes

Processes that are spontaneous in one direction are

non-spontaneous in the reverse direction. The external

conditions often

Spontaneous Processes

• Processes that are spontaneous at one

temperature may be nonspontaneous at other temperatures.

• Above 0 C it is spontaneous for ice to melt.

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Big Idea #5

• The laws of thermodynamics describe the essential role of energy and explain and predict the direction of changes in matter.

Reversible Processes

Increasing the

temperature by an

infinitely small amount causes heat to flow out of the system.

The process can be

reversed by lowering the temp. of the

system.

The system and its

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Irreversible Processes

• Remove partition: gas expands to fill cylinder.

• In order to compress the gas, work has to be done on the system. • Irreversible processes can technically be undone by reversing the

change to the system, but the surroundings will have changed.

Entropy

• _{Entropy (S) is a term coined by Rudolph} Clausius in the 19th century.

• Clausius was convinced of the significance of the ratio of heat

delivered and the temperature at which it is delivered, .q

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Entropy

• Entropy can be thought of as a measure of the randomness of a system.

Entropy

• Like total energy, E, and enthalpy, H, entropy is a state function.

• Therefore,

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Entropy

For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the

process were reversible divided by the temperature: • Remember: a reversible process is one that is

technically impossible. S is a state function. We are interested in the S, so we pretend

S = qrev

Second Law of Thermodynamics

The second law of thermodynamics states that the entropy of the universe increases for spontaneous processes. The entropy of the universe does not change for reversible

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Second Law of Thermodynamics

In other words:

For reversible processes:

S_univ = S_system + S_surroundings = 0

For irreversible processes:

Second Law of Thermodynamics

• An irreversible process increases the entropy of the universe.

– However, because entropy is a state function, the change in entropy of a system is the same

whether the process is reversible or irreversible.

• The second law of thermodynamics can be used to determine whether a process is

reversible or not.

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Recap

Irreversible Processes Reversible Processes

Increase entropy of the universe Cannot change the total entropy of the universe

Returning to original state will require work which will increase entropy of universe

Consider a carnot-cycle-after 1 cycle the system and surroundings are in exact same conditions

Spontaneous- once started, will proceed without outside intervention

Non-spontaneous

Real processes are irrerversible Reversible processes not really possible because of energy “leakage”

Entropy on the Molecular Scale

• Ludwig Boltzmann described the concept of entropy on the molecular level.

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Entropy on the Molecular Scale

• Molecules exhibit several types of motion:

– Translational: Movement of the entire molecule from one place to another.

– Vibrational: Periodic mtion of atoms within a molecule. – Rotational: Rotation of the molecule on about an axis or

Entropy on the Molecular Scale

• Boltzmann envisioned the motions of a sample of molecules at a particular instant in time.

– This would be akin to taking a snapshot of all the molecules.

• He referred to this sampling as a microstate of the

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Entropy on the Molecular Scale

• Each thermodynamic state has a specific number of

microstates, W, associated with it.

• Entropy is

S = k lnW

Entropy on the Molecular Scale

• The change in entropy for a process, then, is

S = k lnW_final  k lnW_initial

W_final W_initial S = k ln

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Entropy on the Molecular Scale

• The number of microstates and,

therefore, the entropy tends to increase with increases in

– Temperature. – Volume.

– The number of

Sample Exercise 19.2 Calculating ΔS for a Phase Change

The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is –38.9 °C, and its molar enthalpy of fusion is ΔHfusion = 2.29 kJ/mol. What is the entropy change of the system

when 50.0 g of Hg(l) freezes at the normal freezing point?.

Solution

Check: The entropy change is negative because heat flows from the system, making qrev negative.

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Entropy and Probability

• Imagine a two-bulbed flask w. 2 gas molecules in one side.

• There are more possible combinations (probability) for molecules to NOT be in the bulb on the right.

Predicting Entropy:

Physical States

• Entropy increases with the freedom of motion of molecules.

• Therefore,

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Solutions

Generally, when a solid is

Predicting Entropy Changes

• In general, entropy increases when

– Gases are formed from liquids and solids;

– Liquids or solutions are formed from solids;

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Sample Exercise 19.3 Predicting the Sign of ΔS

Third Law of Thermodynamics

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Standard Entropies

• These are molar entropy values of substances in their standard states.

• Standard entropies tend to increase with

Standard Entropies

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Entropy Changes

Entropy changes for a reaction can be

estimated in a manner analogous to that by which H is estimated:





n



m

Sample Exercise 19.5 Calculating ΔS from Tabulated Entropies

Calculate ΔSº for the synthesis of ammonia from N₂(g) and H₂(g) at 298 K: N₂(g) + 3 H₂(g) → 2 NH₃(g)

Using the standard entropies in Appendix C, calculate the standard entropy change, ΔS°, for the following reaction at 298 K:

Al₂O₃(s) + 3 H₂(g) → 2 Al(s) + 3 H₂O(g)

Answers: 180.39 J/K

Practice Exercise Solution

Analyze: We are asked to calculate the entropy change for the synthesis of NH₃(g) from its constituent elements.

Plan: We can make this calculation using Equation 19.8 and the standard molar entropy values for the reactants and the products that are given in Table 19.2 and in Appendix C.

Solve: Using Equation 19.8, we have

Substituting the appropriate S° values from Table 19.2 yields

ΔS° = 2S°(NH₃) - [S°(N₂) + 3S°(H₂)]

ΔS° = (2 mol)(192.5 J/mol-K) - [(1 mol)(191.5 J/mol-K) + (3 mol)(130.6 J/mol-K)] = -198.3 J/K

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Entropy Changes in Surroundings

• Heat that flows into or out of the

system changes the entropy of the

surroundings.

• For an isothermal process:

S_surr = qsys

T

• At constant pressure, q_sys is simply

Entropy Change in the Universe

• The universe is composed of the system and the surroundings.

• Therefore,

S_universe = S_system + S_surroundings

• For spontaneous processes

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Gibbs Free Energy

• Since S_surroundings =

and q_system = H_system

This becomes:

S_universe = S_system +

Multiplying both sides by T, we get



T



S

=



H



T



S

H_system T

Gibbs Free Energy

 TS_universe is defined as the Gibbs free energy, G.

• When S_universe is positive, G is negative.

• Therefore, when G is negative, a process is spontaneous.S

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Sample Exercise 19.6 Calculating Free-Energy Change from ΔH°, T, ΔS°

Calculate the standard free energy change for the formation of NO(g) from N₂(g) and O₂(g) at 298 K: N₂(g) + O₂(g) → 2 NO(g)

given that ΔH° = 180.7 kJ and ΔS° = 24.7 J/K. Is the reaction spontaneous under these circumstances?

A particular reaction has ΔH° = 24.6 kJ and ΔS° = 132 J/K at 298 K. Calculate ΔG°. Is the reaction spontaneous under these conditions?

Answers: ΔG° = –14.7 kJ; the reaction is spontaneous.

Practice Exercise Solution

Analyze: We are asked to calculate ΔG° for the indicated reaction (given ΔH°, ΔS° and T) and to predict whether the reaction is spontaneous under standard conditions at 298 K.

Plan: To calculate ΔG°, we use Equation 19.12, ΔG° = ΔH° – T ΔS°. To determine whether the reaction is spontaneous under standard conditions, we look at the sign of ΔG°.

Solve:

Because ΔG° is positive, the reaction is not spontaneous under standard conditions at 298 K.

Gibbs Free Energy

1. If G is negative, the forward reaction is spontaneous.

2. If G is 0, the system

is at equilibrium.

3. If G is positive, the reaction is

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Standard Free Energy Changes

Analogous to standard enthalpies of

formation are standard free energies of formation, G.

f

G =





f f

Free Energy Changes

At temperatures other than 25°C,

G° = H  TS

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Free Energy and Temperature

• There are two parts to the free energy equation:

 H— the enthalpy term

– TS — the entropy term

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Free Energy and Equilibrium

Under any conditions, standard or

nonstandard, the free energy change can be found this way:

G = G + RT lnQ

(Under standard conditions, all concentrations are 1 M,

Free Energy and Equilibrium

• At equilibrium, Q = K, and G = 0. • The equation becomes

0 = G + RT lnK

• Rearranging, this becomes

G = RT lnK

or,

K = e

-G

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Free Energy and Work

• ΔG = max work available from a system.

• But: amount of work obtained is always less than ΔG • Consider a car battery-starter-alternator system

• Some energy lost to non-useful heat through friction, etc.

• Entropy of universe has increased; (Second Law) • Henry Bent: First Law-You can’t win, you can only

break even

Energy Shortage

• As we use energy, we degrade its usefulness

• Consider gasoline: